Massachusetts Institute of Technology
Department of Electrical Engineering & Computer
Science6.041/6.431: Probabilistic Systems Analysis
(a) Consider the derivation of the Markov inequality on page 265
of the text. For the inequalityto be tight, we require that E[Ya] =
E[X]. Since Ya X, this can happen only if Ya = X.(To see this,
consider E[X Ya]. It is a weighted sum of the values that X Ya can
take,where the weights are the corresponding probabilities, and
therefore at least one weight ispositive. But XYa can take only
non-negative values, since Ya X. If this weighted sumhas to be 0,
then X Ya cannot take any positive value, and must be 0.)This
implies that
X = Ya =
{0, if X < a,a, if X a.
This simply means X takes only two values: 0 and a. Now, E[X] =
aP (X = a) = . Hence,P (X = a) = a . This gives the followi ng PMF
for X:
pX(x) =
{1 a , x = 0a , x = a
(b) As shown in pages 266-267 of the textbook, the Chebyshev
bound for a random variable Yis derived by applying the Markov bou
nd for the random variable Z , (Y Y )2. Hence,for the Chebyshev
bound to be tight for Y , the Markov bound has to be tight for Z.
Wenow use the result of part (a). Since E[Z] = E[(Y Y )2] is given
to be 2Y , this means Zmust have the following PMF:
pZ(z) =
{1 2Y
b2, z = 0
2Yb2 , z = b
2
If Z = 0, then Y = Y and this happens with probability 1 2Yb2 .
However, if Z = b
2, thenY can take the va lue of (Y + b) or (Y b). For the mean
to be Y , both these valueshave to be equally likely. This gives
the following PMF for Y :
pY (y) =
1 2Y
b2y = Y
122Yb2, y = (Y b)
122Yb2, y = (Y + b)
(c) Markov upper bound: For z > E[Z] = 10,
P (Z z) E[Z]z
=10
z
Chebyshev upper bound:
P (Z z) = P (Z 10 z 10) P (|Z 10| z 10) 2Z
(z 10)2 =4
(z 10)2
where the first inequality follows from the fact that the event
{Z10 z10} is a subset ofthe event {|Z 10| z 10}, and the second
inequality is an application of the Chebyshevbound.
The figure shows the plot of both bounds as a function of z.
